Transcript Failure of fiber reinforced orthotropic layers.

Failure of fiber reinforced orthotropic
layers.
• Failure often controlled
by the complex
interaction of fibers and
matrix.
• Still, the ideal was to
come up with failure
criteria that are small
generalization of those
used for isotropic
materials.
(a,b) fiber microbuckling; (c) kink band;
(d) shear failure on 45-deg plane.
How many experiments?
• Unlike isotropic materials properties are
different in fiber and matrix directions.
• Brittleness and matrix-fiber interaction mean
that properties are different in tension,
compression and shear.
• Consequently at least five failure stresses
𝑋𝑡 , 𝑋𝑐 , 𝑌𝑡 , 𝑌𝑐 , 𝑆
Max stress and max strain criteria
• Max stress (note that these are no longer
principal stresses)
 X c  1  X t
Yc   2  Yt
 12  S
• Max strain
 X c / E1  1  X t / E1  1t
Yc / E2   2  Yt / E2   2t
 12  S / G12   12s
Example 6.2.1
• Boron Epoxy with 𝐸1 = 204, 𝐸2 = 18.5, 𝐺12 =
5.59𝐺𝑃𝑎, 𝜈12 = 0.23, 𝑡𝑝𝑙𝑦 = 0.125𝑚𝑚.
• Strengths: 𝑋𝑡 = 1260, 𝑋𝑐 = 2500, 𝑌𝑡 =
61, 𝑌𝑐 = 202, 𝑆 = 67𝑀𝑃𝑎.
• What order of magnitude strains can it take?
• Calculate maximum 𝜎𝑥 for varying ply angles.
• Then consider also cases with 𝜎𝑦 = 𝑟𝜎𝑥 , for
r=0.05, and r=0.15.
Solution procedure
• Transform loading stresses 𝜎𝑥 and 𝜎𝑦 to
material coordinates get
 1  (m 2  rn 2 ) x
m  cos 
 2  (n 2  rm 2 ) x
 12  (r  1)mn x
n  sin 
• Then compare each to the limit and find the
maximum value.
Tensile strength
• 𝑋𝑡 = 1260, 𝑋𝑐 = 2500, 𝑌𝑡 =
61, 𝑌𝑐 = 202, 𝑆 = 67𝑀𝑃𝑎.
• Need to look at failure
envelope.
• Does the envelope make
sense?
• Anything surprising?
 1  m 2 x
m  cos 
 2  n 2 x
 12  mn x
n  sin 
Comparing tensile to compressive
strength
• What changed about the envelope?
Biaxial loading
• Anything remarkable?
Tsai-Hill Criterion
• Rodney Hill (1921-2011) Cambridge Prof.
• Hill sought quadratic criterion for anisotropic
materials with equal tensile and compression
propertiesof the form
2
F ( 1   2 ) 2  G ( 2   3 ) 2  H ( 3   1 ) 2  L 23
 M  132  N 122  1
• Stephen Tsai, Stanford Prof. applied to
composites.
• You find coefficients by applying five test
conditions . For example, tensile test in fiber
direction gives 𝐹 + 𝐻)𝑋𝑡2 = 1
• Altogether get
         
2
2
2

    2   1
X
Xt
 S 
 t   Yt 
1
2
1
2
12
Hoffman criterion
• Oscar Hoffman, Lockheed 1967
• To account for difference in tension and
compression need to have linear terms
2
 1
1 1
  12 
1 









 1 
 2   1
X t X c X t X c YtYc  X t X c 
Y
Y
 S 
c 
 t
 12
1  2
 22
• Derivation as homework.
• Criterion is no longer homogeneous. What does
that mean?
• Implication for safety factors. Have a paper on
that with A. Groenwold of U. Stellenbosch.